Question: Is ${751519}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {751519}= &&{7}\cdot100000+ \\&&{5}\cdot10000+ \\&&{1}\cdot1000+ \\&&{5}\cdot100+ \\&&{1}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {751519}= &&{7}(99999+1)+ \\&&{5}(9999+1)+ \\&&{1}(999+1)+ \\&&{5}(99+1)+ \\&&{1}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {751519}= &&\gray{7\cdot99999}+ \\&&\gray{5\cdot9999}+ \\&&\gray{1\cdot999}+ \\&&\gray{5\cdot99}+ \\&&\gray{1\cdot9}+ \\&& {7}+{5}+{1}+{5}+{1}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${751519}$ is divisible by $3$ if ${ 7}+{5}+{1}+{5}+{1}+{9}$ is divisible by $3$ Add the digits of ${751519}$ $ {7}+{5}+{1}+{5}+{1}+{9} = {28} $ If ${28}$ is divisible by $3$ , then ${751519}$ must also be divisible by $3$ ${28}$ is not divisible by $3$, therefore ${751519}$ must not be divisible by $3$.